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Question Number 119483 by Ar Brandon last updated on 24/Oct/20

$$\mathrm{Let}a>0\mathrm{and}f:\mathbb{R}\to \mathbb{R}\mathrm{a}\mathrm{function}\mathrm{satisfying}$$ $$f(\mathrm{x}+a)=1+{[2-3f\left(\mathrm{x}\right)+3f{\left(\mathrm{x}\right)}^2-f{\left(\mathrm{x}\right)}^3]}^{1/3}$$ $$\mathrm{for}\mathrm{all}\mathrm{x}\in \mathbb{R}.\mathrm{Then}\mathrm{a}\mathrm{period}\mathrm{of}f\left(\mathrm{x}\right)\mathrm{is}ka\mathrm{where}k\mathrm{is}$$ $$\mathrm{a}\mathrm{positive}\mathrm{integer}\mathrm{whose}\mathrm{value}\mathrm{is}$$ $$\left(\mathrm{A}\right)1\left(\mathrm{B}\right)2\left(\mathrm{C}\right)3\left(\mathrm{D}\right)4$$

Answered by Olaf last updated on 25/Oct/20

$$f(x+a)=1+{[2-3f\left(x\right)+3f{\left(x\right)}^2-f{\left(x\right)}^3]}^{1/3}$$ $${[f(x+a)-1]}^3=2-3f\left(x\right)+3f{\left(x\right)}^2-f{\left(x\right)}^3$$ $${[f(x+a)-1]}^3=1+[1-3f\left(x\right)+3f{\left(x\right)}^2-f{\left(x\right)}^3]$$ $${[f(x+a)-1]}^3=1-{[f\left(x\right)-1]}^3$$ $$$$ $${[f(x+2a)-1]}^3=1-{[f(x+a)-1]}^3$$ $${[f(x+2a)-1]}^3=1-(1-{[f\left(x\right)-1]}^3)$$ $${[f(x+2a)-1]}^3={[f\left(x\right)-1]}^3$$ $$f(x+2a)-1=f\left(x\right)-1$$ $$f(x+2a)=f\left(x\right)$$ $$\Rightarrow \mathrm{period}\mathrm{is}2a(k=2)$$ $$$$

Commented byAr Brandon last updated on 25/Oct/20

Merci monsieur

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